I've been asked to verify the following equation (I'll explain my notation after a few things):

dW/dt = [a1(t) + b2(t)]*W

I am given that:

W(t) = det[[x1(t), x2(t)][y1(t), y2(t)]]

and

dx/dt = a1(t)*x + b1(t)*y
dy/dt = a2(t)*x + b2(t)*y

My notation is meant to imply the following:

dW/dt, dx/dt, and dy/dt represent the derivatives of the wronskian, x, and y with respect to t.

a1(t), a2(t), b1(t), and b2(t) are functions of the variable t, the numbers are meant as subscripts.

W(t) follows the general format of the wronskian, which is the determinant of the matrix of solutions. In conjunction with that, it is known that

x = c1*x1(t) + c2*x2(t)
y = c1*y1(t) + c2*y2(t)

is a general solution to a system of two homogeneous differential equations.

So, again, I'm interested in verifying that dW/dt = [a1(t) + b2(t)]*W, which the book claims is a simple calculation. I've tried expanding the wronskian, W, in terms of x1(t), x2(t), y1(t), and y2(t), then multiplying by a1(t) and b2(t) as given in the equation I listed above, but that just gets messy, and doesn't seem to match any form of an answer I get for dW/dt in terms of variables.

Any help is most appreciated.

Dec 3rd 2006, 06:16 PM

ThePerfectHacker

I think you are referring to a wonderful theorem, "Abel's Theorem".

Consider,
Where, are some functions defined on an open interval.

Assume, are two function that satisfy the differencial equation throughtout this open interval.
That means,
Now,
Now the Wronskian is differenciable on this open interval and,
The trick is to multiply the system of differencial equations by certain functions that will transform them into the Wronskian form.
Multiply the first equation by :
Multiply the second equation by :
Add these equations, Hacker style,
Thus,
This is a first order linear differencial equation which can be solved easily if is countinous throughout the interval.

I can solve it for thee but I believe thou can complete it thyself.

Dec 3rd 2006, 07:54 PM

primasapere

I can't say that the theorem you mention doesn't look like it should be helpful, but i'm getting stuck where you define the wronskian.

In the section I'm reviewing, the wronskian is defined using elements from the general solution to the system of two differential equations.

If a given homogeneous system has two solutions

x = x1(t)
y = y1(t)

and

x = x2(t)
y = y2(t)

then they define the wronskian as

W(t) = x1(t)y2(t) - x2(t)y1(t)

Should I be reasoning this to be the same as your definition? You define it in terms of y1, y2, while I'm seeing a definition in terms of t.

If it helps, the section is discussing a system of two first order equations in two unknown functions, of the form

dx/dt = F(t,x,y)
dy/dt = G(t,x,y)

Dec 3rd 2006, 08:00 PM

ThePerfectHacker

Quote:

Originally Posted by primasapere

I can't say that the theorem you mention doesn't look like it should be helpful,

What is that supposed to me:mad: . Abel's theorem leads to a theorem that is the Wronskian is zero at a point then it must be zero throughout the interval.

Quote:

If a given homogeneous system has two solutions

x = x1(t)
y = y1(t)

and

x = x2(t)
y = y2(t)

then they define the wronskian as

W(t) = x1(t)y2(t) - x2(t)y1(t)

Should I be reasoning this to be the same as your definition? You define it in terms of y1, y2, while I'm seeing a definition in terms of t.

If it helps, the section is discussing a system of two first order equations in two unknown functions, of the form

dx/dt = F(t,x,y)
dy/dt = G(t,x,y)

I think that is a regular determinant for a 2x2 linear system equatio. Not the Wronskian.

Dec 4th 2006, 12:54 AM

primasapere

I didn't mean to imply an insult or anything. I just wasn't sure I could understand/use your idea.

But after looking for a while, Abel's Theorem, as far as you mention it, relates to the Wronskian in a manner congruent to how my textbook is asking this question, so I think I can manage from here on.